Answer : The value of
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)

conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction
.

![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)

Now we have to calculate the Gibbs free energy of reaction
.
As we know that,

At room temperature, the temperature is 500 K.


Therefore, the value of
for the reaction is -959.1 kJ
Answer:
Some metals can be extracted from compounds by heating with carbon atom because they are less reactive than carbon and some metals cannot be extracted because they are more reactive than carbon atom.
Explanation: If the metal is less reactive than carbon atom so the carbon atoms make bond with oxide or other atom present with metal and the metal is free from that oxide or that element. But if the reactivity of metal is higher than carbon is unable to remove the oxide or element.
Weathering would be chemical because it's more natural than physical
Answer:
Ag+(aq) + Cl-(aq) —> AgCl(s)
Explanation:
2AgNO3(aq) + CaCl2(aq) —>2AgCl(s) + Ca(NO3)2(aq)
The balanced net ionic equation for the reaction above can be obtained as follow:
AgNO3(aq) and CaCl2(aq) will dissociate in solution as follow:
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
CaCl2(aq) —> Ca2+(aq) + 2Cl-(aq)
AgNO3(aq) + CaCl2(aq) –>
2Ag+(aq) + 2NO3-(aq) + Ca2+(aq) + 2Cl-(aq) —> 2AgCl(s) + Ca2+(aq) + 2NO3-(aq)
Cancel out the spectator ions i.e Ca2+(aq) and 2NO3- to obtain the net ionic equation.
2Ag+(aq) + 2Cl-(aq) —> 2AgCl(s)
Divide through by 2
Ag+(aq) + Cl-(aq) —> AgCl(s)
The, the net ionic equation is
Ag+(aq) + Cl-(aq) —> AgCl(s)
Answer:
17.6 moles of oxygen gas.
Explanation:
STP Variables:
P=1 atm
R=0.082
T=273 K
Use the PV=nRT, then plug in.
n=PV/RT
n= (1 atm)(395 L)/(0.082) (273 K)
Simplify.
n=17.6 moles of O2
Hoped this helped.